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Question Number 136243 by Khalmohmmad last updated on 19/Mar/21

lim_(x→0) ((sinx−x)/(tanx−x))=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}{x}−{x}}{\mathrm{tan}{x}−{x}}=? \\ $$

Answered by EDWIN88 last updated on 20/Mar/21

 lim_(x→0)  ((sin x−x)/(tan x−x)) = lim_(x→0)  ((cos x−1)/(sec^2 x−1))  = lim_(x→0)  (((cos x−1)/(1−cos^2 x))).lim_(x→0)  cos x  = −1. lim_(x→0)  ((cos x−1)/((cos x−1)(cos x+1)))  = −(1/2)

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{x}}{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{1}}{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\right).\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\:\mathrm{x} \\ $$$$=\:−\mathrm{1}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{1}}{\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)\left(\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by liberty last updated on 20/Mar/21

 lim_(x→0)  ((sin x−x)/(tan x−x))  = lim_(x→0)  ((x−(x^3 /6)−x)/(x+(x^3 /3)−x)) =− (1/2)

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}−{x}}{\mathrm{tan}\:{x}−{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−{x}}{{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}}\:=−\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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